Outgoing Wave Conditions in Photonic Crystals and ...€¦ · Known radiation conditions Bloch wave...

Negative refractionRadiation conditions

Uniqueness and transmission properties

Outgoing Wave Conditions in PhotonicCrystals and Transmission Properties at

Interfaces

Workshop: “Waves in periodic media and metamaterials”

Agnes Lamacz & Ben Schweizer

November 23, 2016

1 / 29



Description of lightMaxwell’s equations and negatve indexAnother mechanism of negative refraction

Geometric optics vs. Wave equation

Fermat’s principle ofthe fastest path:

Light finds thefastest way to reachthe destination!

sin Θ1sin Θ2

=v1v2

=n2n1

Huygens’ principleof superpositions

Wave equation

∂2t u = ∆u

Numerical solution

2 / 29




Geometric optics vs. Wave equation

Fermat’s principle ofthe fastest path:

Light finds thefastest way to reachthe destination!

sin Θ1sin Θ2

=v1v2

=n2n1

Huygens’ principleof superpositions

Wave equation

∂2t u = ∆u

Numerical solution

3 / 29




Maxwell’s equations and negatve index

Maxwell’s Equations (1865)

curl E = iωµH

curl H = −iωεE

E: electric field, H: magnetic field

H,E ∼ e−iωt

• Re ε < 0 possible• µ is always 1• Reµε < 0: medium is opaque

Veselago (1968)

Materials with negative index

ε < 0 and µ < 0 ⇒ negative index!

Solutions for positive and negative index

Pendry et al. (∼ 2000)Design of a negative index meta-materialUse split rings and wires

4 / 29




Microscopic split-ring geometry

The material parameter εη is

εη =

1 + iκ

η3in the rings

1 else

Effective coefficients µeffand εeff.

The parameter η appears 4×:1 size of the microstructure (η)

2 thin rings (2βη2)

3 very thin slit (2αη4)

4 high conductivity (κη−3)

5 / 29




Microscopic geometry with wires

(Hη, Eη) solves Maxwell, (Hη, Eη)→ (Ĥ, Ê) “geometrically”

Effective Maxwell system (A.Lamacz & B.S., SIAM J.Math.Anal. 2017)

curl Ê = iωµeff Ĥ

curl Ĥ = −iωεeff Êwith negative (for appropriate geometry and Re(εw) < 0) coefficients

µeff = µeff,R = (M̂)−1 and εeff = εeff,R +πγ

2 εW .

6 / 29




An interesting observation about wave transmission

Image taken from

C. Luo, S. G. Johnson, J. D. Joannopoulos, and

J. B. Pendry. All-angle negative refraction

without negative effective index. Phys. Rev.

B, 65:201104, May 2002

Our question:

Is this refraction at a negative indexmaterial?

Explanation of the effect in [LJJP]:The solution in the photonic crystal is aBloch wave, determined by two facts:

it has the right frequency

it conserves the vertical wavenumber

Together, this can explain negativerefraction

7 / 29




An interesting observation about wave transmission

Image taken from

C. Luo, S. G. Johnson, J. D. Joannopoulos, and

J. B. Pendry. All-angle negative refraction

without negative effective index. Phys. Rev.

B, 65:201104, May 2002

Our question:

Is this refraction at a negative indexmaterial?

Explanation of the effect in [LJJP]:The solution in the photonic crystal is aBloch wave, determined by two facts:

it has the right frequency

it conserves the vertical wavenumber

Together, this can explain negativerefraction

8 / 29



Known radiation conditionsBloch wave analysisOutgoing wave condition

This talk

x

x

1

2

Mathematical subject: Radiation condition in periodic media

Homogenious media (Sommerfeld, 1912)

Periodic media (Fliss and Joly, ARMA 2016)

Periodic media with an interface (Lamacz and S., 2016)

9 / 29




Radiation for homogeneous media: Sommerfeld, 1912

Homogeneous problem −∆u = ω2u in Rn

Fundamental solutions

Two fundamental Helmholtz solutions for x ∈ R3:

uout(x) :=1

|x|eiω|x| and uin(x) :=

1

|x|e−iω|x|

Time-dependence e−iωt implies: uout is anoutgoing wave, uin an incoming wave.

Sommerfeld condition

|x|(n−1)/2(∂|x|u− iωu)(x)→ 0 as |x| → ∞(1)

Both elementary solutions decay for

|x| → ∞. It is not reasonable to

demand only a decay property

uout satisfies (1),it is admissible

uin does notsatisfy (1), it isnot admissible

Justification (Sommerfeld): Radiation condition implies uniqueness

10 / 29




Radiation in a periodic wave-guide: Fliss and Joly, 2016

Periodic wave-guide−∇ · (a∇u) = ω2u

Image taken from S. Fliss and P. Joly. Solutions of

the time-harmonic wave equation in periodic

waveguides: asymptotic behaviour and radiation

condition. Arch. Ration. Mech. Anal., 219, 2016

The periodicwaveguide is

neither 2-dimensional (no decay of waves)nor 1-dimensional (variations in vertical direction)

Idea: The solution consists of finitely many outgoing Bloch waves at +∞

Definition (Outgoing radiation condition, Fliss and Joly, 2016)

A function u solves the outgoing radiation condition to the right if

u(.+ (p, 0)) =

N(ω)∑m=1

u+mΦmeipξ+m + w+(.+ (p, 0)), (2)

where w+ has exponential decay at +∞.

Justification: Radiation condition implies existence and uniqueness11 / 29




Radiation in media with an interface

x

x

1

2

The geometry of the transmission problem. We are interestedin waves that are generated in the photonic crystal.

−∇ · (a∇u) = ω2u

Program:

1 Develop an “outgoing wave condition” in a photonic crystal

2 Derive a uniqueness result (justification of the condition)

3 Conclude properties of transmitted waves

12 / 29




Bloch expansion (... on one page!)

1.) f : Rn → R is writtenwith a Fourier transform:

f(x) =

∫Rnf̂(ξ)e2πiξ·x dξ

2.) ξ is written as ξ = k + j withk ∈ Zn and j ∈ [0, 1)n =: Z

f(x) =

∫Z

∑k

f̂(k + j)e2πik·x︸︷︷︸=:F

e2πij·x dj

3.) Periodic F = F (x; j) is expanded in periodic eigenfunctions Ψj,m(x):

F (x; j) =∑m∈N

αj,mΨj,m(x)Uj,m(x) := Ψj,m(x)e

2πij·x solves

−∇ · (a(x)∇Uj,m(x)) = µj,m Uj,m(x)

Result: The operator L0 = −∇ · (a(.)∇) acts as a multiplier:

f(x) =

∫Z

∑m∈N

αj,mUj,m(x) dj, L0f =∫Z

∑m∈N

αj,mµj,mUj,m(x) dj

13 / 29






f(x) =



f(x) =

∫Z

∑k

f̂(k + j)e2πik·x︸︷︷︸=:F

e2πij·x dj


F (x; j) =∑m∈N


2πij·x solves

−∇ · (a(x)∇Uj,m(x)) = µj,m Uj,m(x)


f(x) =

∫Z

∑m∈N


∑m∈N


14 / 29






f(x) =



f(x) =

∫Z

∑k

f̂(k + j)e2πik·x︸︷︷︸=:F

e2πij·x dj


F (x; j) =∑m∈N


2πij·x solves

−∇ · (a(x)∇Uj,m(x)) = µj,m Uj,m(x)


f(x) =

∫Z

∑m∈N


∑m∈N


15 / 29






f(x) =



f(x) =

∫Z

∑k

f̂(k + j)e2πik·x︸︷︷︸=:F

e2πij·x dj


F (x; j) =∑m∈N


2πij·x solves

−∇ · (a(x)∇Uj,m(x)) = µj,m Uj,m(x)


f(x) =

∫Z

∑m∈N


∑m∈N


16 / 29




Expansion of solutions

x

x

1

2

We consider u only on themarked square

After a shift:u ∈ L2((0, Rε)× (0, Rε))

Wave-vector: j ∈ Z := [0, 1)2. Eigenvalue number: m ∈ N0Multiindex: λ = (j,m) ∈ IK . Basis: U+λ (x) := Ψ

+λ (x) e

2πiθ(λ)·x/ε

u(x) =∑λ∈IK

α+λU+λ (x)

Expansion of an arbitrary functionu in Bloch waves

Idea: For “outgoing solutions” we demand:

u (on the right) consists only of right-going Bloch modes

Note: u periodic −→ Bloch expansion is a sum

17 / 29




Energy transport of Bloch waves

In the expansion u(x) =∑λ∈IK

α+λU+λ (x) which Bloch modes are outgoing?

Recall: The Poynting vector P := E ×H measures the energy flux

Poynting number

For λ ∈ I, the Poynting number P+λ describes the right-going energy:

P+λ := Im−∫Yε

Ū+λ (x) e1 ·[aε(x)∇U+λ (x)

]dx

Index sets: Left-going waves and “vertical waves”

I+




Outgoing wave condition

Given u on R× (0, h), height h = εK, factor R ∈ KNLarge square: RYε = (0, Rε)× (0, Rε)ũ : R2 → C the h-periodic vertical extensionDefine u+R : RYε → C by

u+R(x1, x2) := ũ(Rε+ x1, x2)

Expand u+R:

u+R(x) =∑λ∈IR

α+λ,RU+λ (x)

The coefficients (α+λ,R)λ∈I encodethe behavior of u for large x1

Definition (Outgoing wave condition)

We say that u satisfies the outgoing wave condition on the right if:

a)∫ h0

∫ L+1L|u|2 is bounded, independently of L ≥ 0, and b)

−∫RYε

∣∣Π+




Our wish-list

Transmission problem

a constant on the left, periodic on the right

Helmholtz equation: −∇ · (a∇u) = ω2u, periodic in vertical direction

Outgoing wave conditions, on the right:

−∫RYε

∣∣Π+ 0

There exists a solution to the problem

The solution to the problem is unique

Uniqueness cannot be expected

There are surface-waves −→ no uniqueness! S. Bozhevolnyi/Aalborg Univ.

21 / 29




Our wish-list

Transmission problem

a constant on the left, periodic on the right

Helmholtz equation: −∇ · (a∇u) = ω2u, periodic in vertical direction

Outgoing wave conditions, on the right:

−∫RYε

∣∣Π+ 0

There exists a solution to the problem

The solution to the problem is unique

Uniqueness cannot be expected

There are surface-waves −→ no uniqueness! S. Bozhevolnyi/Aalborg Univ.22 / 29



Bloch measures and uniquenessTransmission condition and outlook

Bloch measures

G. Allaire and C. Conca. Bloch wave homogenization and spectral asymptotic analysis. J. Math. Pures Appl. 1998

Let uR ∈ L2(WR;C) be a sequence

uR(x) =∑λ∈IR

α±λU±λ (x)

Discrete Bloch-measure for fixed l ∈ N0:

ν±l,R :=∑

λ=(j,l)∈IR

|α±λ |2 δj

where δj denotes the Dirac measure in j ∈ Z.If, as R→∞,

ν±l,R → ν±l,∞

in the sense of measures, then

ν±l,∞ ∈M(Z) is a Bloch measure generated by u

The Brillouin zone Z = [0, 1)2.

A periodic u is expanded with

discrete values of j ∈ Z.

23 / 29




Uniqueness result

Frequency assumption with Bloch-eigenvalues µ±m(j):

ω2 < infj∈Z,m≥1

µ+m(j)

Theorem (Uniqueness)

Let u and ũ be two solutions of the transmission problem and letv := u− ũ be the difference. Then:

the Bloch measure of v is supported on vertical waves

for non-singular frequencies ω: the Bloch measure of v vanishes

Interpretation of the result:Waves must be localizedto the interfaceFor general ω: verticalwaves are possible

Figure: The indices

j ∈ Z corresponding to“vertical waves”

24 / 29




Uniqueness follows from energy conservation

Poynting vector bilinear form b±R : L2(WR;C)×H1(WR;C)→ C:

b+R(u, v) := −∫WR

ū(x) e1 · [a(x)∇v(x)] dx

Let v solve the Helmholtz equation with coefficients a, use

ϑ(x) :=

1 if |x1| ≤ εR2− |x1|εR if εR < |x1| < 2εR0 if |x1| ≥ 2εR x1

and the test-function ϑ(x) v(x):∫R

∫ h0

{aϑ |∇v|2 + a ∂x1ϑ v ∂x1v

}= ω2

∫R

∫ h0

ϑ |v|2

Take the imaginary parts and obtain the energy conservation

Im b−R(v−R , v

−R

)= Im b+R

(v+R , v

+R

)Result: If both terms have opposite sign, they must vanish! 25 / 29




Show ν±l,∞ = 0 for l ≥ 1

Let δ > 0 be a number with δ ≤ |ω2 − µ(j,l)|2 for all j and l ≥ 1. Then,formally,

δ −∫WR

∣∣∣Πev,+l≥1 (u+R )∣∣∣2 = δ ∑λ=(j,l)∈IR

l≥1

∣∣〈u+R , Uλ〉R∣∣2≤

∑λ=(j,l)∈IR

l≥1

∣∣(ω2 − µλ)〈u+R , Uλ〉R∣∣2≤∑λ∈IR

∣∣〈ω2u+R , Uλ〉R − 〈µλu+R , Uλ〉R∣∣2(∗)=∑λ∈IR

∣∣〈L0(u+R) , Uλ〉R − 〈µλu+R , Uλ〉R∣∣2 = 0The calculation can be made precise with cut-off functions on largesquares. Result for Bloch measure: ν±l,∞ = 0 for l ≥ 1Similar calculations yield:supp(ν±0,∞) ⊂

{j ∈ Z |µ±0 (j) = ω2 , j2 ∈ N/K

}26 / 29




Transmission condition

Assume again: Frequency below second bandThe vertical wave number is conserved:

Theorem (Transmission conditions)

Let u be a solution of the transmission problem.Let ν±l,∞ be a Bloch measure to u.

Then: ν±l,∞ = 0 for l ≥ 1. For l = 0 holds

supp(ν±0,∞) ⊂{j ∈ Z |µ±0 (j) = ω2 , j2 ∈ N/K

}and

supp(ν±0,∞) ⊂ {j ∈ Z | j2 = k2} ∪ J±=0,0

Waves must have:

the correct energyand

the correct k2 (or bevertical)

The theorem follows from uniqueness: Compare u with its projection tothe vertical wave number k2

27 / 29




Conclusions and open problems

Corollary: For non-singular frequencies ω, the Bloch measure of u issupported on {j ∈ Z | j2 = k2}.

−→ conservation of the vertical wave numberNegative refraction can therefore be explained ...

... using that the vertical wave number is conserved.

Open for the transmission problem:

1 Existence with limiting absorption?

2 Vertical waves excluded?

3 Implementation of the outgoing wavecondition?

Thank you!

28 / 29




Conclusions and open problems

Corollary: For non-singular frequencies ω, the Bloch measure of u issupported on {j ∈ Z | j2 = k2}.

−→ conservation of the vertical wave numberNegative refraction can therefore be explained ...

... using that the vertical wave number is conserved.

Open for the transmission problem:

1 Existence with limiting absorption?

2 Vertical waves excluded?

3 Implementation of the outgoing wavecondition?

Thank you!

29 / 29

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